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Advocard [28]
3 years ago
6

Dianne's teacher has a chunk of dry ice for the class to look at. Dry ice is a kind of frozen gas—it is very cold. The teacher w

arns her class not to touch it. When Dianne holds her hand near it, it feels like the dry ice is radiating cold. What is really happening?
A) Heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it.
B) Chemicals in the dry ice are tricking Diane's nerves into thinking she feels cold.
C) Cold energy from the dry ice is being transferred to Diane's hand even though she is not touching it.
D) Small pieces of the dry ice are breaking off and hitting Diane's hand, making it feel cold.
Physics
2 answers:
Gelneren [198K]3 years ago
4 0
The answer is A. Hope this helps you with your work.
Lera25 [3.4K]3 years ago
3 0

Answer:

A) Heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it.

Explanation:

In this example, we learn that Dianne feels cold when she holds her hand close to the dry ice. In fact, Dianne feels as it the cold is moving from the dry ice to her hand. However, this is not what is happening. Changes in temperature occur when heat is transferred from a warm body to a colder one. Therefore, what is actually occurring is that heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it, and this makes her feel cold.

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Colonist who belived in complete​
velikii [3]

Answer:

Natural right

Explanation:

That  is all what i know

5 0
3 years ago
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m
Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

\omega = \sqrt{\frac{k}{m}}

here we know

m = 3.5 kg

k = 270 N/m

now we have

\omega = \sqrt{\frac{270}{3.5}}

\omega = 8.78 rad/s

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

v = \omega \sqrt{A^2 - x^2}

0.55 = 8.78 \sqrt{A^2 - 0.020^2}

A = 0.066 m

Part b)

Maximum speed of SHM at its mean position is given as

v_{max} = A\omega

v_{max} = 0.066(8.78) = 0.58 m/s

4 0
3 years ago
How can submarines use echolocation to tell how close they are to the bottom of the ocean?
Lapatulllka [165]
To locate a specific target or to determine how close submarines are to the seafloor, they use active and passive sound navigation and ranging (or a SONAR, in simple terms.) It emits pulses of sound waves that travel through the water, reflect off the target and relayed back to the ship. By determining how fast the sound wave travels back, the computers on the sub calculate how far they are from the target.

Hope this helps. 
3 0
3 years ago
Read 2 more answers
What term refers to a universal fact sometimes based on mathematical equations?
noname [10]

Answer:

Scientific law

Explanation:

3 0
2 years ago
A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite
julia-pushkina [17]

Answer:

<em>a. Angle= 28.82°</em>

<em>b. Approved. He will get cold but he should be able to make it across</em>

Explanation:

Velocity Vector

The velocity is a physical quantity that measures how fast or slow at a particular direction some object is moving. It must be expressed as a vector with both a magnitude and direction. If the object is confined to move in one direction, then we can use the speed as the scalar (magnitude only) equivalent of the velocity.

a.

The explorer wants to swim across a river to his campsite, as shown in the image below. The river has a velocity vr and the explorer can swim at ve in still water. If he swam directly to the campsite, he would end up in a point below it because the river would push him down. He must swim with a velocity such that he overcomes the stream but he advances to its objective. Let's call the angle he must swim at respect to the shoreline to achieve his goal. The explorer's velocity can be decomposed in its rectangular components vx and vy. To overcome the river's velocity:

v_{ey}=v_r

We can compute the vertical component of the explorer's velocity as

v_{ey}=|v_e|cos\alpha

Thus

v_r=|v_e|cos\alpha

Solving for \alpha

\displaystyle cos\alpha=\frac{v_r}{|v_e|}

\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876

Then we have the angle is

\alpha=28.82^o

b.

The horizontal component of the explorer's velocity is

v_{ex}=0.759sin28.82^o

v_{ex}=0.366\ m/s

This is the real velocity the explorer is having directly to the campsite

Knowing that

\displaystyle v=\frac{x}{t}

Solving for t

\displaystyle t=\frac{x}{v}

Calculating the time it takes the explorer to cross the river

\displaystyle t=\frac{29.3}{0.366}

t=80\ sec

Since this value is less than the limit value of hypothermia (300 sec), the decision is

Approved. He will get cold but he should be able to make it across

3 0
3 years ago
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